Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
Previous << 1 .. 352 353 354 355 356 357 < 358 > 359 360 361 362 363 364 .. 486 >> Next

(s+1/2)2- 9/4
3 2
to be the same as that found above.
f(t) = (2/3)u2(t)e (t2)/2 sinh—(t-2), which can be shown
22
By completing the square in the denominator of F we can write F(s) = (2s+1)/[(2s+1) + 4]. This has the form
G(2s+1) where G(u) = u/(u +4). We must find ?-1{G(2s+1)}. Applying the results of Problem 19(c), we
have ?-1{F(s)} = —e-t/2 cos(-t).
22
If the approach of Problem 21 is used we find
f(t) = (1/3)e2t/3 sinh(t/3), which is equivalent to the given answer using the definition of sinht.
Assuming that term-by-term integration of the infinite series is permissible and recalling that ?{uc(t)} = e-cs/s
for s > 0, we have ?{f(t)} = (1/s) + (-1)k ?{uk(t)}
k=1
• •
= (1/s) + ^^(-1)k e-ks/s = ?(-0 k]/s. We recognize k=1 k=0
the last infinite series as the geometric series, ^ark,
k=0
with a = 1 and r = -e s. This series converges to [1/(1+e-s) ] if |r| < 1. Hence,
?{f(t)} = (1/s)[1/(1+e-s)], s > 0.
Section 6.4
111
28. Using the definition of the Laplace transform we have
F(s) = ?{f(t)} = J0 e-stf(t)dt. Since f is periodic with period T, we have f(t+T) = f(t). This suggests that we
f -st
rewrite the improper integral as I e f(t)dt =
^ Mn+1)T_st
y InT e f(t)dt. The periodicity of f also suggests
n=0
that we make the change of variable t = r + nT. Hence,
“ T “ T
F(s) = Jo e-s(r+nT)f(r+nT)dr = (e sT) n J0 e-rsf(r)dr,
n=0 n=0
where we have used the fact that
f(r+nT) = f(r+(n-1)T) = ... = f(r+T) = f(r) from the definition that f is periodic. We recognize this last

series as the geometric series, aun, with
n=0
T
-rs -sT
a = |0e f(r)dr and u = e . The geometric series
converges to a/(1-u) for |u| < 1 and consequently we obtain
-sT -1 T -rs
F(s) = (1 - e-sT)-1 |o e-rsf(r)dr, s > 0.
30. The function f is periodic with period 2. The result of
2 -st -2s
Problem 28 gives us ?{f(t)} = I e^fdddt/d-e- s). Calculating the integral we have
•2 i*1 i*2
J0 e stf(t)dt = J0 e stdt - J1 e stdt
2s -s
= d-e^/s + (e s-es)/s = (e^^e^+D/s
-s 2
= (1-e ) /s. Since the denominator of ?{f(t)}, 1 - e-2s, may be written as d-e^d+e-6) we obtain the desired answer.
Section 6.4, Page 321
1. f(t) can be written in the form f(t) = 1 - up/2(t) and thus the Laplace transforms of the D.E. is (s2+1)Y(s) - sy(0) - y'(0) = (1/s) - e-Ps/2/s. Introducing the I.C. and solving for Y(s), we obtain
Y(s) = (s2+1) -1 + [s(s2+1)] -1 - e-Ps/2/s(s2+1). Using partial fractions on the second and third terms we find
112
Section 6.4
Y(s) = (1/s) + (s2+1) -1 - s/(s2+1) - e-Ps/2/s + e-Ps/2s/(s2+1). The inverse transform of the first three terms can be obtained directly from Table 6.2.1. Using Theorem 6.3.1 to find the inverse transform of the last two terms we
have ?- 1{e-ps /2/s} = up/2(t)g(t - p/2) where g(t) = ?-1{1/s} = 1 and
?-1{e Ps/2s/(s2+1)} = up/2(t)h(t - p/2) where
h(t) = ?-1{s/(s2+1)} = cost. Hence, y = 1 + sint - cost + up/2(t)[cos(t - p/2) - 1]
= 1 + sint - cost - up/2(t)[1 - sint]. The graph of the forcing function is a unit pulse for 0 ? t < p/2 and 0 thereafter. The graph of the solution will be composed of two segments. The first, for 0 ? t < p/2, is a sinusoid oscillating about 1, which represents the system response to a unit forcing function and the given initial conditions. For t > p/2, the forcing function, f(t), is zero and the "initial" conditions are y(p/2) = lim 1 + sint - cost = 2 and
t—nr/2
y'(p/2) = lim cost + sint = 1. In this case the system
t—nr/2
response is y(t) = 2sint - cost, which is a sinusoid oscillating about zero.
3. According to Theorem 6.3.1,
?{u2p(t)sin(t-2p)} = e-2ps ?{sint} = e-2ps/(s2+1). Transforming the D.E., we have
(s2+4)Y(s) - sy(0) - y'(0) = 1/(s2+1) - e-2ps/(s2+1). Introducing the I.C. and solving for Y(s), we obtain -2 ps 2 2
Y(s) = (1-e s/(s +1)(s +4). We apply partial fractions to write
x i- 2 - -1 i 2 „ _1 -2ps 2 - -1 -2ps ^
Y(s) = [s +1) - (s +4) - e (s +1) + e (s +4) ]/3.
We compute the inverse transform of the first two terms directly from Table 6.2.1 after noting that
2 -1 2 (s +4) = (1/2)[2/(s +4)]. We apply Theorem 6.3.1 to the
last two terms to obtain the solution,
y =(1/3){sint-(1/2)sin2t-u2p(t)[sin(t-2p)-(1/2)sin2(t-2p)]}.
This may be simplified, using trigonometric identities, to y = [(2sint - sin2t)(1-u2p(t))]/6. Note that the forcing function is sint - sin(t-2p) = 0 for t > 2p. The solution is y(t) = 2sint - sin2t for 0 ? t < 2p. Thus
y(2p ) = 0 and y'(2p ) = 2cos2p - 2cos4p = 0. Hence the "initial" value problem for t > 2p is y" + 4y = 0, y(2p) = 0, y'(2p) = 0, which has the trivial solution y = 0.
Section 6.4
113
Taking the Laplace transform, applying the I.C. and using Theorem 6.3.1 we have (s2+s + 5/4)Y(s) = (1-e-ps/2)/s2. Thus
1-e-s/2
Y(s) =
22 s (s +s+5/4)
= e-«s/)i/5 16/25 + (16/25)s-4/25 1
^ s2 s (s + 1/2) 2+1
-ps/2
= (1-e 'lH(s), where we have used partial fractions and completed the square in the denominator of the last term. Since the numerator of the last term of H
16
can be written as —[(s+1/2) - 3/4], we see that
Previous << 1 .. 352 353 354 355 356 357 < 358 > 359 360 361 362 363 364 .. 486 >> Next